an/bn=n/3^nTn=1/3^1+2/3^2+3/3^3+……+(n-2)/3^(n-2)+(n-1)/3^(n-1)+n/3^n3Tn=1/3^2+2/3^3+3/3^4+……+(n-2)/3^(n-1)+(n-1)/3^n+n/3^(n+1)相减:
设数列{an/bn}的前n项和为Tn,试比较4Tn/3与(2n^2+3n-2)*2^(n-1)的大小 an=n,
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